I sat down to write a linear algebra take-home exam problem where I would give a $4\times4$ matrix $A$ and ask for bases for these six spaces: $\mathrm{Col}\,A$, $\mathrm{Row}\,A$, $\mathrm{Nul}\,A$, $\mathrm{Col}\,A^t$, $\mathrm{Row}\,A^t$, $\mathrm{Nul}\,A^t$. Trivially, a basis for $\mathrm{Col}\,A$ will work for $\mathrm{Row}\,A^t$, and likewise with $\mathrm{Row}\,A$ and $\mathrm{Col}\,A^t$; that was meant to be part of the test.
I expected $\mathrm{Nul}\,A$ and $\mathrm{Nul}\,A^t$ to be different. I worked on the first $4\times4$ matrix that came to mind: $\begin{bmatrix}1&2&3&4\\5&6&7&8\\9&10&11&12\\13&14&15&16\end{bmatrix}$. To my surprise, the standard method for finding the basis of the null space led me to see that in this case, $\mathrm{Nul}\,A=\mathrm{Nul}\,A^t$. (And since I do not want students to think this will always be the case, I have to try a different matrix.)
As a consequence of the orthogonality of row and null spaces, $\mathrm{Row}\,A=\mathrm{Row}\,A^t$. So we have a matrix where: $$\mathrm{Col}\,A^t=\mathrm{Row}\,A=\mathrm{Row}\,A^t=\mathrm{Col}\,A$$ $$\mathrm{Nul}\; A=\mathrm{Nul}\;A^t$$ all of which is equivalent to just knowing $$\mathrm{Row}\, A=\mathrm{Col}\,A$$
or to just knowing $$\mathrm{Nul}\, A=\mathrm{Nul}\,A^t$$And these are equivalent to the existence of a matrix $B$ such that $AB=A^t$, since the rows of $A$ (columns of $A^t$) are linear combinations of the columns of $A$.
Now any invertible matrix would automatically have these properties, since the null space of an invertible matrix is $\{\vec0\}$. And any symmetric matrix will automatically have these properties, since $A=A^t$. But it seems like a noninvertible, asymmetric square matrix will rarely have these properties. (In the $2\times 2$ case, there are none.)
With general $n\times n$ matrices again, I would like to know
- Is there a name for this kind of matrix ($\mathrm{Row}\,A=\mathrm{Row}\,A^t$)? Perhaps additionally demanding noninvertible asymmetric specimens?
- Does this kind of matrix have any interesting applications?
- If we fix $\mathrm{rank}\, A$, do matrices of this form make an interesting surface in $M_{n\times n}$?
- If so, what is the dimension of this surface?
Best Answer
A geometric characterization of these matrices would be the following:
An $(n\times n)$-matrix $A$ is the same as a linear transformation $A:\ {\mathbb R}^n\to{\mathbb R}^n$. Provide ${\mathbb R}^n$ with the standard scalar product.
Claim: The condition ${\rm Col}\,A={\rm Row}\,A$ is equivalent to ${\rm ker}\,A=({\rm im}\, A)^\perp$. In particular any orthogonal projection would have this property.
Proof. The condition ${\rm Col}\,A={\rm Row}\,A$ means that ${\rm im}\,A={\rm im}\,A^t$. Take an $x\in{\ker}\,A$. Then $$0=\langle A x,y\rangle=\langle x,A^t y\rangle\qquad\forall y\ ;$$ therefore $x$ is orthogonal to ${\rm im}\,A^t={\rm im}\,A$. It follows that ${\rm ker}\,A\subset({\rm im}\,A)^\perp$, whence ${\rm ker}\,A=({\rm im}\, A)^\perp$ by counting dimensions.
Conversely, assume that ${\rm ker}\,A=({\rm im}\, A)^\perp$. Take an $x\in{\rm im}\,A^t$. Then $x=Az$ for a $z\in{\mathbb R}^n$, and we have $$\langle u,x\rangle=\langle u,A^t z\rangle=\langle Au,z\rangle =0\qquad\forall u\in{\rm ker}\,A\ .$$ It follows that ${\rm im}\,A^t\subset({\rm ker}\,A)^\perp={\rm im}\,A$ and therefore ${\rm im}\,A^t={\rm im}\,A$, as $A$ and $A^t$ have the same rank.